Question

Answer this Question of iit jee .
Chapter → Electrostatics​

Answer 1

EXPLANATION.

Three charges are placed at the vertices of equilateral triangle.

Charges 'q' on each.

Sides of equilateral triangle = l cm.

As we know that,

Potential energy = (U).

⇒ U = kq₁q₂/r.

k = 1/4πεο.

⇒ A = kq²/l. - - - - - (1).

⇒ B = kq²/l. - - - - - (2).

⇒ C = kq²/l. - - - - - (3).

⇒ U = A + B + C.

⇒ U = kq²/l + kq²/l + kq²/l.

⇒ U = 3kq²/l.

⇒ U = 3q²/4πεο.l

⇒ U = 1/4πεο. 3q²/l.

Option [C] is the correct answer.

Answer 2

Explanation:

We know ,

$potential \: energy \: ,U = \frac{kq1q2}{ {r}^{2} }$

So,

Net Potential energy,

$U = U1 + U2 + U3$

$→U = \frac{k {q}^{2} }{l} + \frac{k {q}^{2} }{l} + \frac{k {q}^{2} }{l} \\ = \frac{3k {q}^{2} }{l} \\ = \frac{1}{4\piε₀} \frac{3 {q}^{2} }{l}$